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- anonymous

What is a quartic function with only the two real zeroes given?
x = –1 and x = –3

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- anonymous

What is a quartic function with only the two real zeroes given?
x = –1 and x = –3

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- anonymous

You can go with (x + 1)^2 times (x + 3)^2
The reason this works is that it will give you a x^4 term and you still have only the 2 roots.
Roots are of the form:
(x - root) for you factors.

- anonymous

So, basically you have both roots twice, so that's why each factor has a 2 on it for an exponent.

- anonymous

You could also have put a 3 on one of them and left the other root as a single power (no exponent)

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- anonymous

Is this making sense to you?

- anonymous

I am lost completely

- anonymous

np, I'll give you a little more detail:
First off, a quartic polynomial function has as its highest exponent, an x^4 term. That means that if you have an equation of the form:
y = (x - root1)(x - root2)(x - root3)(x - root4)
you will have quartic function. But we can have only 2 different roots, so a good choice is to have root1 twice and root 2 twice:
y = [(x - root1)^2][(x - root2)^2]

- anonymous

That's the same as:
y = (x - root1)(x - root1)(x - root2)(x - root2)
which also has root1 twice and root2 twice. It's just writing out the factors without the exponents. This equation will give you 0 for "y" if you replace x with either root, which is what you want to solve this problem.

- anonymous

And to finish this off, you are given x = -1 and x = -3 as the only 2 roots you can have. And since the factor to give "0" using the root is:
(x - root)
one factor is [x - (-1)] which is (x + 1)
the other is [x - (-3)] which is (x + 3)
So, your polynomial is:
y = (x + 1)(x + 1)(x + 3)(x + 3)

- anonymous

You have other choices, but the above polynomial works. You could also have:
y = (x + 1)(x + 3)(x + 3)(x + 3)
or
y = (x + 1)(x + 1)(x + 1)(x + 3)
Both of those are quartics having only 2 different roots.
That should make it all clearer now. Now are you good on this?

- anonymous

Here's a graph of the first solution

- anonymous

sorta so which one would it be?
y = x4 – 4x3 – 4x2 – 4x – 3
y = –x4 + 4x3 + 4x2 + 4x + 3
y = x4 + 4x3 + 3x2 + 4x – 4
y = x4 + 4x3 + 4x2 + 4x +3
I understand how you did that i just don't know which one it will be.

- anonymous

- anonymous

I didn't know that you had choices. That changes the whole way you approach this problem. You can solve it 2 ways that I know of. The first way rst and the best and fastest is to perfrom synthetic division on each choice until you hit upon the choice that has only those 2 roots.

- anonymous

Number #1 is out because -1 is not a root. So, I eliminated that one so far.

- anonymous

Number 2 is out because it also has +1 as a root.

- anonymous

Are you sure you wrote those choices down correctly?

- anonymous

I actually copied and paste so yeah

- anonymous

And the roots are supposed to be -1 and -3 ?

- anonymous

yeah that is why i couldn't figure it out. That is what confused me

- anonymous

It looks like like the last one works. I tried it again.

- anonymous

okay thanks again

- anonymous

you're welcome.

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